Geometric Brownian Motion With Time-Dependent Coefficients

Why Time-Dependent Coefficients Change the Formula

In geometric Brownian motion, the process is multiplied by its current value in both the drift and diffusion terms. When the coefficients vary deterministically with time, the same exponential-transform idea still works, but each coefficient contributes through an integral. The volatility contributes twice: its stochastic integral determines the random fluctuation, while its squared integral supplies the Ito correction in the drift of the exponent.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) with respect to the filtration. Fix a finite horizon \(T\). Assume that \(\mu:[0,T]\to\mathbb{R}\) is deterministic, measurable, and integrable, and that \(\sigma:[0,T]\to\mathbb{R}\) is deterministic, measurable, and square-integrable. Thus \(\int_0^T|\mu(u)|\,du<\infty\) and \(\int_0^T\sigma(u)^2\,du<\infty\). Let \(S_0\in\mathbb{R}\) be deterministic.

The \(dB_t\) integral is an Ito integral, not a Stratonovich integral. Ito’s formula uses \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). In particular, converting a Stratonovich equation to Ito form can add a drift correction; the equation studied here is already in Ito form.

The Explicit Solution

Proof. Define \(A_t=\int_0^t(\mu(u)-\tfrac12\sigma(u)^2)\,du\) and \(Y_t=\int_0^t\sigma(u)\,dB_u\). The deterministic integrand \(\sigma\) is progressively measurable and satisfies \(\mathbb{E}\int_0^T\sigma(u)^2\,du<\infty\), so \(Y_t\) is a well-defined Ito integral. Set \(S_t=S_0e^{A_t+Y_t}\). For \(f(t,y)=S_0e^{A_t+y}\), the derivatives, at almost every \(t\), are \(f_t=(\mu(t)-\tfrac12\sigma(t)^2)f\), \(f_y=f\), and \(f_{yy}=f\). Also \(dY_t=\sigma(t)\,dB_t\) and \(d[Y]_t=\sigma(t)^2\,dt\). Ito’s formula, including its second-derivative correction, gives $$ dS_t=f_t\,dt+f_y\,dY_t+\frac12f_{yy}\,d[Y]_t =\left(\mu(t)-\frac12\sigma(t)^2\right)S_t\,dt+\sigma(t)S_t\,dB_t+\frac12\sigma(t)^2S_t\,dt =\mu(t)S_t\,dt+\sigma(t)S_t\,dB_t. $$ This also holds when \(S_0=0\), since then \(f\) and all its derivatives are zero.

To check that the stochastic integral in this equation is in \(\mathcal{L}^2\), write \(v(0,t)=\int_0^t\sigma(u)^2\,du\). The deterministic Brownian integral \(Y_t\) is normal with mean zero and variance \(v(0,t)\), so the normal moment-generating function gives $$ \mathbb{E}[S_t^2]=S_0^2\exp\left(2\int_0^t\mu(u)\,du+v(0,t)\right). $$ The right-hand side is bounded on \([0,T]\), since \(\mu\) is integrable and \(\sigma^2\) is integrable. Consequently \(\mathbb{E}\int_0^T\sigma(t)^2S_t^2\,dt<\infty\). Also \(\mathbb{E}\int_0^T|\mu(t)S_t|\,dt<\infty\), by the bound on \(\mathbb{E}[S_t^2]\), the Cauchy–Schwarz inequality, and integrability of \(\mu\). Thus the displayed differential equation gives the required integral equation.

For uniqueness, let \(\widetilde S_t\) be any other continuous adapted solution with the same initial value. Define \(R_t=\exp(-A_t-Y_t)\). Applying Ito’s formula to the exponential with exponent \(-A_t-Y_t\) gives \(dR_t=(-\mu(t)+\sigma(t)^2)R_t\,dt-\sigma(t)R_t\,dB_t\). The Ito product rule then yields $$ d(\widetilde S_tR_t)=R_t\,d\widetilde S_t+\widetilde S_t\,dR_t+d[\widetilde S,R]_t=0, $$ because the drift coefficients sum to \(\mu+(-\mu+\sigma^2)-\sigma^2=0\), and the diffusion coefficients sum to \(\sigma-\sigma=0\). Therefore \(\widetilde S_tR_t=\widetilde S_0R_0=S_0\). Since \(R_t>0\), this forces \(\widetilde S_t=S_0/R_t=S_0e^{A_t+Y_t}\), proving uniqueness. The product calculation is valid locally for semimartingales; the identity holds on the full interval by localization. \(\square\)

Conditional Law from Time \(s\) to Time \(t\)

For \(0\le s\le t\le T\), define $$ m(s,t)=\int_s^t\left(\mu(u)-\frac12\sigma(u)^2\right)\,du,\qquad v(s,t)=\int_s^t\sigma(u)^2\,du. $$ The quantity \(v(s,t)\) is nonnegative and may equal zero. Subtracting the exponents in the explicit solution gives the factorization $$ S_t=S_s\exp\left(m(s,t)+\int_s^t\sigma(u)\,dB_u\right). $$ This identity follows by splitting each integral at \(s\); it does not require division by \(S_s\), so it remains valid when \(S_s=0\).

Proof. For a deterministic step function on \([s,t]\), its Brownian integral is a finite sum of Brownian increments after time \(s\). Those increments are independent of \(\mathcal{F}_s\), and the sum is normal with mean zero and variance equal to the integral of the squared step function. Approximate \(\sigma\mathbf{1}_{[s,t]}\) in \(L^2([0,T])\) by deterministic step functions \(h_n\). By the Ito isometry, \(\int_s^t h_n(u)\,dB_u\) converges in \(L^2(\mathbb{P})\) to \(I=\int_s^t\sigma(u)\,dB_u\), and \(\int_s^t h_n(u)^2\,du\to v(s,t)\). For any bounded \(\mathcal{F}_s\)-measurable random variable \(U\) and real \(q\), \(L^2\) convergence implies \(L^1\) convergence, and \(|e^{iqx}-e^{iqy}|\le |q||x-y|\). Passing to the limit in the step-function identity therefore gives $$ \mathbb{E}[Ue^{iqI}]=\mathbb{E}[U]e^{-q^2v(s,t)/2}. $$ This identity states that \(I\) is independent of \(\mathcal{F}_s\) and has characteristic function \(e^{-q^2v(s,t)/2}\), that of \(N(0,v(s,t))\); when \(v(s,t)=0\), it is the constant zero. Applying this fact to the factorization of \(S_t\), with \(S_s\) measurable with respect to \(\mathcal{F}_s\), proves the stated conditional law, including the case \(S_s=0\). \(\square\)

Degenerate Cases and a Common Correction Error

When \(v(s,t)=0\), the volatility vanishes almost everywhere on \([s,t]\), so the stochastic integral over that interval is zero. The conditional transition is then deterministic: \(S_t=S_se^{m(s,t)}\). A zero current value also stays zero, because the explicit formula multiplies the exponential by \(S_s\). Neither case requires taking a logarithm of the process.

A common error is to omit the Ito correction or to count the diffusion coefficient twice. For example, take constant \(\mu=1\), \(\sigma=2\), and the incorrect candidate \(G_t=e^{t+2B_t}\). With \(f(t,y)=e^{t+2y}\), the derivatives are \(f_t=f\), \(f_y=2f\), and \(f_{yy}=4f\). Since \(y=B_t\), Ito’s correction is \(\tfrac12 f_{yy}\,dt=2f\,dt\), not \(\tfrac12(4f)(2^2)\,dt\). Therefore $$ dG_t=3G_t\,dt+2G_t\,dB_t, $$ so \(G_t\) has drift \(3G_t\), not the required \(G_t\). The correct exponent is \((1-\tfrac12\cdot 2^2)t+2B_t=-t+2B_t\). For \(F_t=e^{-t+2B_t}\), the derivatives with respect to \(t\) and \(B_t\) are \(-F_t\), \(2F_t\), and \(4F_t\); the drift is \(-F_t+\tfrac12(4F_t)=F_t\), as required.

The key distinction is that \(f_{yy}\) above is already the second derivative with respect to the Brownian coordinate \(y=B_t\). If instead one uses a coordinate \(y=2B_t\), then its quadratic variation is \(4\,dt\), and the second derivative must be taken with respect to that coordinate. Mixing the two choices double-counts the diffusion factor.